Faculty of Social Sciences Induction Block: Maths

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Faculty of Social Sciences Induction Block
Maths Statistics Lecture 4

• Probability, Randomness and Different Types of
  Events
• Dr Gwilym Pryce

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Plan

• 1. Introduction
• 2. Randomness Probability
• 3. Complementary and Disjoint Events
• 4. When 2 events can occur together
• 5. Independent events
• 6. Contingent events

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Birthday Prediction

• I predict that there are between one and two
  birthdays in the class this week
• How did I do this?...

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• A/ 1 in 52 chance that it is your birthday
• 70 people in the room
• expected number of birthdays this week 70
  1/52 1.346
• Q/ Does that mean that every class of 70 has
  1.346 birthdays that week?
• A/ No. It means that in the long run (ie lots of
  lecture classes of size 70) the average number of
  people with birthdays will 1.346

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e.g. Birthdays in many classes
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• it can take a long time for the long run average
  to emerge

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Coin tossing example
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2. Randomness and Probability

• A phenomenon is random if individual outcomes are
  uncertain but there is nonetheless a regular
  distribution of outcomes in a large number of
  repetitions.

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• The probability of any outcome of a random
  variable is the proportion of times the outcome
  would occur in a very long series of repetitions.
  
• I.e. long-term relative frequency number of
  times an event occurs in the long run divided by
  the number of possible outcomes.

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Probability of an event

• Probability that a strangers birthday is this
  week 1/52
• Probability that flipped coin is heads 1/2
• Probability of picking a red ball from a bag of 2
  red and 8 blue is
• 0.2 or 20 or 1 in 5 chance or 5 to 1
  against.
• NB probabilities always lie between 0 and 1.

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3. Probability of an event not occurring

• This is called the complement of an event
• It is calculated as
• 1 - probability of the event occurring
• P(A) 1 - P(A)

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Disjoint Events

• Two events A and B are disjoint if they have no
  outcomes in common and so can never occur
  simultaneously
• E.g. You have one die want to know the
  probability of rolling a 4 or a 6.
• Answer If the die is fair, the chance of rolling
  a 4 is 1/6. Chance of rolling a 6 is also 1/6.
  Chance of rolling either a 4 or a 6 (cant have
  both at the same time) is 1/6 1/6 1/3.
• More generally, when A and B are disjoint
  (mutually exclusive),
• P(A or B) P(A) P(B)

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Venn diagram of disjoint events
B
A
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4. When 2 events can occur together

• Suppose that you now have a die and a coin
• What is the chance of getting a 4 and/or heads?

A Roll a 4
B Toss heads (!)
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• Probability of rolling 4 1/6
• Probability of tossing heads 1/2
• Probability of rolling 4 and/or heads
• 1/6 1/2 4/6
• Or is it?
• Havent we double counted the probability of
  getting 4 and a heads?
• So we need to deduct this probability (1/61/2)

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What we should really do is take away (1/61/2)

• Probability of rolling 4 and/or heads
• 1/6 1/2 - (1/61/2)
• 4/6 1/12
• 7/12

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To see this, suppose you only have a two sided
die (can only roll a one or a two) and a coin and
want to know Pr(1 or tails or both)

• We know that Pr(1) ½ and Pr(heads) ½
• If we use the incorrect formula we get
• Pr(1 or heads or both) ½½ 1
• Which is clearly incorrect
• There are 4 possible outcomes, 3 of which qualify
  as 1 or tails or both
• (1,H) (1,T) (2,H) (2,T)
• I.e. Pr(1 or heads or both) ¾
• We have double counted Pr(both) Pr(1,H), where
• Pr(1,H) ½ ½ ¼

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• The correct formula is therefore
• P(A or B) P(A) P(B) - P(A and B)
• Probability Notation
• A ? B A union B A or B occur
• A ? B A intersection B A and B occur
• So we can re-write the above as
• P(A ? B) P(A) P(B) - P(A ? B)

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5. Probability of two Independent events occurring

• If knowing that one event occurs does not affect
  the outcome of another event, we say those two
  outcomes are independent.
• And if A and B are independent, and we know the
  probability of each of them occurring, we can
  calculate the probability of them both occurring

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Example 2 sided die and coin, find Pr(1 and H)

• Answer ½ x ½ ¼
• Rule P(A ? B) P(A) x P(B)

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e.g. Tossing one coin twice

• Suppose
• A 1st toss is a head
• B 2nd toss is a head
• what is the probability of A ? B?
• Answer A and B are independent and are not
  disjoint. P(A) 0.5 and P(B) 0.5. P (A ? B)
  0.5 x 0.5 0.25.

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6. Probability of two contingent events occurring

• If knowing that one event occurs does change the
  probability that the other occurs, then two
  events are not independent and are said to be
  contingent upon each other
• If events are contingent then we can say that
  there is some kind of relationship between them
• So testing for contingency is one way of testing
  for a relationship

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Example of contingent events

• There is a 70 chance that a child will go to
  university if its parents are middle class, but
  only a 10 chance if its parents are working
  class. Given that there is a 60 chance of a
  childs parents being working class, what are the
  chances that a child will be working class and go
  to University? What proportion of people at
  university will be from working working class
  backgrounds?

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6 of all children are working class and end up
going to University
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as percent of all children
Working class Middle class
Go to University 6 28
Do not go to University 54 12
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at Uni from WC parents?

• Of all children, only 34 end up at university
  (6 WC 28 MC)
• I.e 6 out of every 34 University students are
  from WC parents
• 6/34 17.65 of University students are WC

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• Probability theory states that
• if x and y are independent, then the probability
  of events x and y simultaneously occurring is
  simply equal to the product of the two events
  occurring
• if x and y are not independent, then
• Prob(x ? y) Prob(x) ? Prob(y given that x has
  occurred)

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Test for independence

• We can use these two rules to test whether events
  are independent
• Does the distribution of observations across
  possible outcomes resemble the random
  distribution we would get if events were
  independent?
• I.e. if we assume independence and calculate the
  expected number of of cases in each category, do
  these figures correspond fairly closely to the
  actual distribution of outcomes found in our data?

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Example 1 Is there a relationship between social
class and education? We might test this by
looking at categories in our data of WC, MC,
University, no University. Suppose we have 300
observations distributed as follows
Working class Middle class
Go to University 18 84
Do not go to University 162 36
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• To do the test for independence we need to
  compare expected with observed.
• How do we calculate ei, the expected number of
  observations in category i?
• I.e. number of cases expected in i assuming that
  the variables are independent
• the formula for ei is the probability of an
  observation falling into category i multiplied
  simply by the total number of observations.
• I.e. No contingency

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• So, if UNIY or UNIN and WC or MC are independent
  (i.e. assuming H0) then
• Prob(UNIY ? WC) Prob(UNIY)?Prob(WC)
• so the expected number of cases for each of the
  four mutually exclusive categories are as
  follows

Working class Middle class
Go to University P(UNIY) x P(WC) x n P(UNIY) x P(MC) x n
Do not go to University P(UNIN) x P(WC) x n P(UNIN) x P(MC) x n
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• But how do we work out
• Prob(UNIY) and Prob(WC)
• which are needed to calcluate Prob(UNIY ? WC)
• Prob(UNIY ? WC) Prob(UNIY)?Prob(WC)
• Answer we assume independence and so estimate
  them from out data by simply dividing the total
  observations by the total number in the given
  category
• E.g. Prob(UNIY) Total no. cases UNIY ? All
  observations
• (18 84) / 300 0.34

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Working class Middle class
Go to University P(UNIY) x P(WC) x n (no. at Uni / n) x (no. WC/n) x n P(UNIY) x P(MC) x n (no. at Uni / n) x (no. MC/n) x n
Do not go to University P(UNIN) x P(WC) x n (no.not Uni / n) x (no. WC/n) x n P(UNIN) x P(MC) x n (no. not Uni / n) x (no. MC/n) x n
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Working class Middle class
Go to University 18 84 102
Do not go to University 162 36 198
180 120 300
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Working class Middle class
Go to University P(UNIY) x P(WC) x n (102 / 300) x (180 /300) x 300 P(UNIY) x P(MC) x n (102 / 300) x (120 /300) x 300
Do not go to University P(UNIN) x P(WC) x n (198 / 300) x (180 /300) x 300 P(UNIN) x P(MC) x n (198 / 300) x (120 /300) x 300
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Expected count in each category
Working class Middle class
Go to University (102 / 300) x (180 /300) x 300 .34 x .6 x 300 61.2 (102 / 300) x (120 /300) x 300 .34 x .4 x 300 40.8
Do not go to University (198 / 300) x (180 /300) x 300 .66 x .6 x 300 118.8 (198 / 300) x (120 /300) x 300 .66 x .4 x 300 79.2
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So we have the actual count (I.e. from our data
set)
Working class Middle class
Go to University 18 84
Do not go to University 162 36
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And the expected count (I.e. the numbers wed
expect if we assume class education to be
independent of each other)
Working class Middle class
Go to University 61.2 40.8
Do not go to University 118.8 79.2
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What does this table tell you?
Working class Middle class
Go to University Actual count 18 84
Expected count 61.2 40.8
Do not go to University Actual count 162 36
Expected count 118.8 79.2
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• It tells you that if class and education were
  indeed independent of each other
• I.e. the outcome of one does not affect the
  chances of outcome of the other
• Then youd expect a lot more working class people
  in the data to have gone to university than
  actually recorded (61 people, rather than 18)
• Conversely, youd expect far fewer middle class
  people to have gone to university (half the
  number actually recorded).

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But remember, all this is based on a sample, not
the entire population

• Q/ Is this discrepancy due to sampling variation
  alone or does it indicate that we must reject the
  assumption of independence?